Complex Numbers — MATHS STD 11 Science — Question

3. -8

Solution:

Using $\text{a}^4+\text{b}^4=(\text{a}^2+\text{b}^2)^2-2\text{a}^2\text{b}^2$

$(1+\text{i})^4+(1-\text{i})^4$

$=\Big((1+\text{i})^2+(1-\text{i})^2\Big)^2-2(1+\text{i})^2(1-\text{i})^2$

$=(1+\text{i}^2+2\text{i}+1+\text{i}^2-2\text{i})^2-2(1+\text{i}^2+2\text{i})(1+\text{i}^2-2\text{i})$

$=(1-1+2\text{i}+1-1-2\text{i})^2-2(1-1+2\text{i})(1-1-2\text{i})$

$=(0)-2(2\text{i})(-2\text{i}) \ (\because\text{i}^2=-1)$

$=8\text{i}^2$

$=-8$​​​​​​​